A wire of natural length 50 cm, diameter 1.5 mm and Young's modulus 3.2 GPa is stretched to a new length of 52.4 cm which is below the limit of proportionality How much work was done in order for this to happen?

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A wire of natural length 50 cm, diameter 1.5 mm and Young's modulus 3.2 GPa is stretched to a new length of 52.4 cm which is below the limit of proportionality How much work was done in order for this to happen?

Answer 1

The work done in stretching the wire was \(3.2542016 \times 10^{3} J\).

Answer 2

## Step1: First, we'll calculate the change in length ΔL. That's the final stretched length of the wire subtracted from its original length. Given that the final length of the wire is \(52.4 \mathrm{~cm}\) and the original length is \(50.0 \mathrm{~cm}\), we also need to convert the lengths from cm to m for calculating the further measurements in standard international units. \[\Delta L = L_f - L_i\] \[\Delta L = 52.4 \mathrm{~cm} - 50 \mathrm{~cm} = 2.4 \mathrm{~cm} = 0.024 \mathrm{m}\] ## Step2: Next, we calculate the 'strain', which is also the elongation, simply the change in length (from Step1: \(0.024 \mathrm{~m}\)) divided by the original length \(L_i\) = \(50 \mathrm{~cm}\) = \(0.5 \mathrm{~m}\)). ### \[\varepsilon = \frac{\Delta L}{L_i}\] ### \[\varepsilon = \frac{0.024 \mathrm{~m}}{0.5 \mathrm{~m}} = 0.048\] ## Step3: Use Young's Modulus (Y) relationship for isotropic & vecrtical materials, where stress / strain = Y: ### \[Y = \frac{\sigma}{\varepsilon}\] Solving for stress (σ), we have: ### \[\sigma = Y \cdot \varepsilon\] Substituting into the equation values for \(Y = 3.2 \mathrm{~GPa} = 3.2 \times 10^9 \mathrm{N/m^2}\) and \(\varepsilon = 0.048\), we get stress \(\sigma = 1.536 \times 10^{8} \mathrm{~N/m^2}\). ## Step4: To calculate the work done to stretch the wire, we leverage the formula: ### \[W = \frac{1}{2} \cdot\sigma\cdot\Delta V\] The volume change \(\Delta V\), is merely the product of cross-sectional area multiplied by the change in length \(\Delta L\). Let's calculate the volume change: Using the given wire diameter \(d = 1.5 \mathrm{~mm} = 0.0015 \mathrm{~m}\), and noting that the radius \(r = d/2\), we substitute \(r = 0.00075 \mathrm{~m}\) into the live formula for the wire/cylinder shape of the wire: (Area =π\(r^{2}\) ) ### \[A = \phi (r^{2}) = 3.1416\times(0.00075 m)^{2} = 1.767 x 10^{-6} m^{2}\] Finally, volume change \(\Delta V = A \bullet \Delta L = 1.767 x 10^{-6} m^{2} \times 0.024 m = 4.241\times 10^{-8} m^{3}\)). Injecting this value into the work formula from before; we'll have \[W = \frac{1}{2} \cdot1.536 \times 10^{8} \mathrm{~N/m^2} \cdot4.241 \times 10^{-8} m^{3} = 3.2542016 \times 10^{3} N \_ball_Calc..\text_to_vector{}\]

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A wire of natural length 50 cm, diameter 1.5 mm and Young's modulus 3.2 GPa is stretched to a new length of 52.4 cm which is below the limit of proportionality How much work was done in order for this to happen?

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